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Step-by-Step Solution
Step 1: Identify the known quantities
• Magnetic moment, M = 9.85 \times 10^{-2}\,\text{A}\cdot \text{m}^2
• Moment of inertia, I = 5 \times 10^{-6}\,\text{kg}\cdot \text{m}^2
• Number of oscillations (n) in total time, n = 10
• Total time for n oscillations, t = 5\,\text{s}
• Hence, the time period for one oscillation, T = \frac{t}{n} = \frac{5}{10} = 0.5\,\text{s} .
Step 2: Write the formula for the time period of oscillation
The time period T of a magnetic needle in a uniform magnetic field B is given by:
T = 2\pi \sqrt{\frac{I}{M\,B}}
Step 3: Rearrange to solve for B
Rearranging the formula for B :
T = 2\pi \sqrt{\frac{I}{M\,B}}
\quad \Longrightarrow \quad
B = \frac{4\pi^2\,I}{M\,T^2}.
Step 4: Substitute the numerical values
Let us substitute the given values into the expression for B :
\begin{aligned}
I &= 5 \times 10^{-6}\,\text{kg}\cdot \text{m}^2, \\
M &= 9.85 \times 10^{-2}\,\text{A}\cdot \text{m}^2, \\
T &= 0.5\,\text{s}, \\
\pi^2 &= 9.85.
\end{aligned}
So,
B = \frac{4 \times \pi^2 \times 5 \times 10^{-6}}
{\,\left(9.85 \times 10^{-2}\right) \times (0.5)^2}.
Step 5: Calculate step by step
1. (0.5)^2 = 0.25 .
2. 4 \times \pi^2 = 4 \times 9.85 = 39.4.
3. Numerator = 39.4 \times (5 \times 10^{-6}) = 1.97 \times 10^{-4}.
4. Denominator = (9.85 \times 10^{-2}) \times 0.25 = 0.0985 \times 0.25 = 0.024625.
5. Thus,
B = \frac{1.97 \times 10^{-4}}{0.024625}
\approx 8 \times 10^{-3} \,\text{T}.
Step 6: Convert Tesla to milliTesla
1\,\text{T} = 1000\,\text{mT} . Hence, 8 \times 10^{-3}\,\text{T} = 8\,\text{mT} .
Final Answer
The magnitude of the uniform magnetic field is 8\,\text{mT} .
